This paper is devoted to prove, in a nonclassical function space,
the weak solvability of parabolic integrodifferential equations with a nonclassical boundary conditions. The investigation is made by means of approximation by the Rothes method which is based on a semidiscretization of the given problem with respect to the time variable.

1. Introduction

The purpose of this paper is to study the solvability of the following equation:
∂v∂t(x,t)-∂2v∂x2(x,t)=∫0ta(t-s)k′(s,v(x,s))ds+g(x,t),(x,t)∈(0,1)×[0,T],
with the initial condition
v(x,0)=V0(x),x∈(0,1),
and the integral conditions
∫01v(x,t)dx=E(t),t∈[0,T],∫01xv(x,t)dx=G(t),t∈[0,T],
where v is an unknown function, E, G, and V0 are given functions supposed to be sufficiently regular, while k′ and a are suitably defined functions satisfying certain conditions to be specified later and T is a positive constant.

Since 1930, various classical types of initial boundary value problems have been investigated by many authors using Rothe time-discretization method; see, for instance, the monographs by Rektorys [1] and Kačur [2] and references cited therein. The linear case of our problem, that is, ∫0ta(t-s)k′(s,v(x,s))ds=0, appears, for instance, in the modelling of the quasistatic flexure of a thermoelastic rod (see [3]) and has been studied, firstly, by the second author with a more general second-order parabolic equation or a 2m-parabolic equation in [3–5] by means of the energy-integrals method and, secondly, by the two authors via the Rothe method [6–8]. For other models, we refer the reader, for instance, to [9–12], and references therein.

The paper is organized as follows. In Section 2, we transform problem (1.1)–(1.3) to an equivalent one with homogeneous integral conditions, namely, problem (2.3). Then, we specify notations and assumptions on data before stating the precise sense of the desired solution. In Section 3, by the Rothe discretization in time method, we construct approximate solutions to problem (2.3). Some a priori estimates for the approximations are derived in Section 4, while Section 5 is devoted to establish the existence and uniqueness of the solution.

2. Preliminaries, Notation, and Main Result

It is convenient at the beginning to reduce problem (1.1)–(1.3) with inhomogeneous integral conditions to an equivalent one with homogeneous conditions. For this, we introduce a new unknown function u by setting

u(x,t)=v(x,t)-R(x,t),(x,t)∈(0,1)×[0,T],
where

R(x,t)=6(2G(t)-E(t))x-2(3G(t)-2E(t)).
Then, the function u is seen to be the solution of the following problem:

∂u∂t(x,t)-∂2u∂x2(x,t)=∫0ta(t-s)k(s,u(x,s))ds+f(x,t),(x,t)∈(0,1)×[0,T],u(x,0)=U0(x),x∈(0,1),∫01u(x,t)dx=0,t∈[0,T],∫01xu(x,t)dx=0,t∈[0,T],
where
f(x,t)=g(x,t)-∂R(x,t)∂t,U0(x)=V0(x)-R(x,0),k(s,u(x,s))=k′(s,u(x,s))-R(x,t).
Hence, instead of looking for the function v, we search for the function u. The solution of problem (1.1)–(1.3) will be simply given by the formula v(x,t)=u(x,t)+R(x,t).

We introduce the function spaces, which we need in our investigation. Let L2(0,1) and L2(0,T;L2(0,1)) be the standard function spaces. We denote by C0(0,1) the linear space of continuous functions with compact support in (0,1). Since such functions are Lebesgue integrable, we can define on C0(0,1) the bilinear form given by
((u,v))=∫01ℑxuℑxvdx,
where

ℑxu=∫0xu(ζ,·)dζ.
The bilinear form (2.5) is considered as a scalar product on C0(0,1) for which C0(0,1) is not complete.

Definition 2.1.

We denote by B21(0,1) a completion of C0(0,1) for the scalar product (2.5), which is denoted by (·,·)B21(0,1), called the Bouziani space or the space of square integrable primitive functions on (0,1). By the norm of function u from B21(0,1), we understand the nonnegative number
∥u∥B21(0,1)=(u,u)B21(0,1)=∥ℑxu∥,where ∥v∥ denotes the norm of v in L2(0,1).

For u∈L2(0,1), we have the elementary inequality
∥u∥B21(0,1)≤12∥u∥.

We denote by L2(0,T;B21(0,1)) the space of functions which are square integrable in the Bochner sense, with the scalar product

Since the space B21(0,1) is a Hilbert space, it can be shown that L2(0,T;B21(0,1)) is a Hilbert space as well. The set of all continuous abstract functions in [0,T] equipped with the norm

sup0≤τ≤T∥u(·,τ)∥B21(0,1)
is denoted C(0,T;B21(0,1)). Let V be the set which we define as follows:

V={v∈L2(0,1);∫01v(x)dx=∫01xv(x)dx=0}.

Since V is the null space of the continuous linear mapping l: L2(0,1)→ℝ2, φ→l(φ)=(∫01φ(x)dx,∫01xφ(x)dx), it is a closed linear subspace of L2(0,1), consequently V is a Hilbert space endowed with the inner product (·,·). Strong or weak convergence is denoted by → or ⇀, respectively. The letter C will stand for a generic positive constant which may be different in the same discussion.

Lemma 2.2 (Gronwall's lemma).

(a1)
Let x(t)≥0, h(t), y(t) be real integrable functions on the interval [a,b]. If
y(t)≤h(t)+∫atx(s)y(s)ds,∀t∈(a,b), then
y(t)≤h(t)+∫ath(s)x(s)exp(∫atx(τ)dτ)ds,∀t∈(0,T).

In particular, if x(t)≡C is a constant and h(t) is nondecreasing, then
y(t)≤h(t)ec(t-a),∀t∈(0,T).

(a2)
Let {ai}ibe a sequence of real nonnegative numbers satisfying
ai≤A+Bh∑k=1i-1ak,∀i=1,2,…,whereA, B, and h are positive constants, such that Bh<1. Then
ai≤Aexp[B(i-1)h],takes place for all i=1,2,….

In the sequel, we make the following assumptions.

(H1) Functions f:[0,T]→L2(0,1) and a:[0,T]→ℝ are Lipschitz continuous, that is,

(H2) The mapping k:[0,T]×V→L2(0,1) is Lipschitz continuous in both variables, that is,

∃l3∈ℝ+;∥k(t,u)-k(t′,u′)∥≤l3[|t-t′|+∥u-u′∥],
for all t,t′∈I, u,u′∈V, and satisfies

∃l4,l5∈ℝ+;∥k(t,u)∥B21(0,1)≤l4∥u∥B21(0,1)+l5,
for all t∈I and all u∈V, where l4 and l5 are positive constants.

(H3)U0∈H2(0,1) and

∫01U0(x)dx=∫01xU0(x)dx=0.

We will be concerned with a weak solution in the following sense.

Definition 2.3.

A function u:I→L2(0,1) is called a weak solution to problem (2.3) if the following conditions are satisfied:

u∈L∞(I,V)∩C(I,B21(0,1)),

u is strongly differentiable a.e. in I and du/dt∈L∞(I,B21(0,1)),

u(0)=U0 in V,

the identity
(dudt(t),v)B21(0,1)+(u(t),v)=(∫0ta(t-s)k(s,u(s))ds,v)B21(0,1)+(f(t),v)B21(0,1)holds for all v∈V and a.e. t∈[0,T].

To close this section, we announce the main result of the paper.

Theorem 2.4.

Under assumptions (H1)–(H3), problem (2.3) admits a unique weak solution u, in the sense of Definition (2.3).

3. Construction of an Approximate Solution

In order to solve problem (2.3) by the Rothe method, we proceed as follows. Let n be a positive integer, we divide the time interval I=[0,T] into n subintervals Ijn:=[tj-1n,tjn], j=1,…,n, where tjn:=jhn and hn:=T/n. Then, for each n≥1, problem (2.3) may be approximated by the following recurrent sequence of time-discretized problems. Successively, for j=1,…,n, we look for functions ujn∈V such that
ujn-uj-1nhn-d2ujndx2=hn∑i=0j-1a(tjn-tin)k(tin,uin)+fjn,∫01ujn(x)dx=0,∫01xujn(x)dx=0,
starting from

u0n=U0,δu0n=d2dx2U0+f(0),
where ujn(x):=u(x,tjn), δujn:=(ujn-uj-1n)/hn, fjn(x):=f(x,tjn). For this, multiplying for all j=1,…,n, (3.1) by ℑx2v:=∫0x(∫0ξv(τ)dτ)dξ and integrating over (0,1), we get

∫01δujn(x)ℑx2vdx-∫01d2ujndx2(x)ℑx2vdx=hn∫01∑i=0j-1a(tjn-tin)k(tin,uin)ℑx2vdx+∫01fjnℑx2vdx.
Noting that, using a standard integration by parts, we have

ℑ12v=∫01(1-ξ)v(ξ)dξ=∫01v(ξ)dξ-∫01ξv(ξ)dξ=0,∀v∈V.
Carrying out some integrations by parts and invoking (3.6), we obtain for each term in (3.5)
∫01δujnℑx2vdx=-(δujn,v)B21(0,1),∫01d2ujndx2(x)ℑx2vdx=(ujn,v),hn∫01∑i=0j-1a(tjn-tin)k(tin,uin(x))ℑx2vdx=-hn∑i=0j-1a(tjn-tin)(k(tin,uin),v)B21(0,1),
and for the last one

∫01fjn(x)ℑx2v(x)dx=-(fjn,v)B21(0,1).
By virtue of (3.7) and (3.8), (3.5) becomes
(δujn,v)B21(0,1)+(ujn,v)=hn∑i=0j-1a(tjn-tin)(k(tin,uin),v)B21(0,1)+(fjn,v)B21(0,1),
or

(ujn,v)B21(0,1)+hn(ujn,v)=hn2∑i=0j-1a(tjn-tin)(k(tin,uin),v)B21(0,1)+hn(fjn,v)B21(0,1)+(uj-1n,v)B21(0,1).
Let η(·,·):V×V→ℝand Lj(·):V→ℝ be two functions defined by

It is easy to see that the bilinear form η(·,·) is continuous on V and V-elliptic, and the form Lj(·) is continuous for each j=1,…,n. Then, Lax-Milgram lemma guarantees the existence and uniqueness of ujn, for all j=1,…,n.

4. A Priori EstimatesLemma 4.1.

There exists C>0 such that, for all n≥1 and all j=1,…,n, the solution uj of the discretized problem (3.1)–(3.4) satisfies the estimates
∥ujn∥≤C,∥δujn∥B21(0,1)≤C.

Proof.

Testing the difference (3.9)j-1-(3.9)j with v=δujn(∈V), taking into account assumptions (H1)–(H3) and the Cauchy-Schwarz inequality, we obtain
∥δujn∥B21(0,1)+∥ujn-uj-1n∥B21(0,1)≤∥δuj-1n∥B21(0,1)+C13hn2∑i=0j-2∥uin∥B21(0,1)+C13hn+C13hn∥uj-1n∥B21(0,1),where
C1:=3max{l2ζ,Tl2ζ+M1ζ+l1},M1:=maxt∈I|a(t)|,ζ:=max{l4,l5}.Multiplying the left-hand side of the last inequality with (1-(C1/3)hn)(<1) and adding the terme
23C1hn[∥ujn-uj-1n∥B21(0,1)-∥δujn∥B21(0,1)](<0),we get
(1-C1hn)[∥δujn∥B21(0,1)+∥ujn∥B21(0,1)]≤[∥uj-1n∥B21(0,1)+∥δuj-1n∥B21(0,1)]+C1hn2∑i=0j-2∥uin∥B21(0,1)+C1hn.
Applying the last inequality recursively, it follows that
(1-C1hn)j[∥δujn∥B21(0,1)+∥ujn∥B21(0,1)]≤[∥u0n∥B21(0,1)+∥δu0n∥B21(0,1)+C1T]+TC1hn∑i=0j-2∥uin∥B21(0,1),
or, by virtue of Lemma 2.2, there exists n0∈ℕ* such that
∥δujn∥B21(0,1)+∥ujn∥B21(0,1)≤C2,∀n≥n0,where
C2:=(exp(TC1)+1)[∥δu0n∥B21(0,1)+∥u0n∥B21(0,1)+TC1]×exp[(exp(TC1)+1)TC1],and so our proof is complete.

We address now the question of convergence and existence.

5. Convergence and Existence

Now let us introduce the Rothe function un(t):I→V obtained from the functions uj by piecewise linear interpolation with respect to time

un(t)=uj-1n+δujn(t-tj-1n),inIjn,
as well the step functions ũn(t), ûn(t), f̃n(t), and k̃(t,ũn(t)) defined as follows:

There exist C>0 such that the estimates
∥un(t)∥≤C,∥ũn(t)∥≤C,∀t∈I,∥dundt(t)∥B21(0,1)≤C,fora.e.t∈I,∥ũn(t)-un(t)∥B21(0,1)≤Chn,∥ûn(t)-un(t)∥B21(0,1)≤Chn,∀t∈I,∥k̃n(t)∥≤C,∀t∈I,
hold for all n∈ℕ*.

Proof.

For the inequalities (5.5), (5.6), and (5.7) see [6, Corollary 4.2.], whereas for the last inequality, assumption (H2) and estimate (4.1) guarantee the desired result.

Proposition 5.2.

The sequence (un)n converges in the norm of the space C(I,B21(0,1)) to some function u∈C(I,B21(0,1)) and the error estimate
∥un-u∥C(I,B21(0,1))≤Chn
takes place for all n≥n0.

Proof.

By virtue of (5.2), (5.3), and (5.4) the variational equation (3.9) may be rewritten in the form
(dundt(t),v)B21(0,1)+(ũn(t),v)=(k̃n(t),v)B21(0,1)+(f̃n(t),v)B21(0,1),
for a.e. t∈[0,T].In view of (5.10), using (5.6) and (5.8) with the fact that
∥f̃n(t)∥B21(0,1)≤M2:=maxt∈I∥f(t)∥B21(0,1)<∞,we obtain
|(ũn(t),v)|≤(∥k̃n(t)∥B21(0,1)+∥f̃n(t)∥B21(0,1)+∥dundt(t)∥B21(0,1))∥v∥B21(0,1)≤C∥v∥B21(0,1),a.e.t∈[0,T].
Now, for n, m being two positive integers, testing the difference (5.10)n-(5.10)m with v=un(t)-um(t) which is in V, with the help of the Cauchy-Schwarz inequality and taking into account that
2(ddtu(t),u(t))B21(0,1)=ddt∥u(t)∥B21(0,1)2,a.e.t∈[0,T],and, by virtue of (5.12) we obtain after some rearrangements
12ddt∥un(t)-um(t)∥B21(0,1)2+∥ũn(t)-ũm(t)∥2≤C∥um(t)-ũm(t)∥B21(0,1)+C∥ũn(t)-un(t)∥B21(0,1)+∥k̃n(t)-k̃m(t)∥B21(0,1)∥un(t)-um(t)∥B21(0,1)+∥f̃n(t)-f̃m(t)∥B21(0,1)∥un(t)-um(t)∥B21(0,1),a.e.t∈[0,T].
To derive the required result, we need to estimate the third and the last terms in the right-hand side, for this, let t be arbitrary but fixed in (0,T], without loss of generality we can suppose that there exist three positive integers p,q and β, such that
t∈(tp-1n,tpn]∩(tq-1m,tqm],n=βm,tpn=tqm.Hence, using (5.4) we can write
∥k̃n(t)-k̃m(t)∥B21(0,1)=hm∥∑j=0p-1[∑i=jββ(j+1)-1(a(tpn-tjn)k(tjn,ujn)-a(tqm-tim)k(tim,uim))]∥B21(0,1).By virtue of assumption (H1) and the fact that |a(tpn-tjn)-a(tqm-tim)|≤Chn, there exist εn∈[0,Chn] such that
∥k̃n(t)-k̃m(t)∥B21(0,1)≤hm∑j=0p-1[∑i=jββ(j+1)-1∥(Chn-εn)k(tjn,ujn)∥B21(0,1)+|a(tqm-tim)|∥k(tjn,ujn)-k(tim,uim)∥B21(0,1)].Therefore, recalling assumptions (H1), (H2) and having in mind that εn∈[0,Chn], we estimate
∥k̃n(t)-k̃m(t)∥B21(0,1)≤hm∑j=0p-1[∑i=jββ(j+1)-1Chn+C(hn+∥ujn-uim∥B21(0,1))],from where, we derive for all s∈(tim,ti+1m]∥k̃n(t)-k̃m(t)∥B21(0,1)≤hm∑j=0p-1[∑i=jββ(j+1)-1Chn+C(hn+∥ũn(s)-un(s)∥B21(0,1)+∥un(s)-um(s)∥B21(0,1)+∥um(s)-ũm(s)∥B21(0,1))∑i=jββ(j+1)-1Chn].Taking the supremum with respect to s from 0 to t in the right-hand side, invoking the fact that s∈(tim,ti+1m]⊂(tj-1n,tjn] and estimate (5.7), we obtain
∥k̃n(t)-k̃m(t)∥B21(0,1)≤hm∑i=0q-1(Chn+Csup0≤s≤t∥un(s)-um(s)∥B21(0,1)),so that
∥k̃n(t)-k̃m(t)∥B21(0,1)≤Chn+Csup0≤s≤t∥un(s)-um(s)∥B21(0,1).
Let t∈(tp-1n,tpn]∩(tq-1m,tqm], from assumption (H1) it follows that
∥f̃n(t)-f̃m(t)∥B21(0,1)=∥f(tpn)-f(tqm)∥B21(0,1)≤l1|tpn-tqm|≤l1hn.
Ignoring the second term in the left-hand side of (5.14) which is clearly positive and using estimates (5.5), (5.7), (5.21), and (5.22) yield
ddt∥un(t)-um(t)∥B21(0,1)2≤C(hn+hm)+Csup0≤s≤t∥un(s)-um(s)∥B21(0,1)2,a.e.t∈[0,T].Integrating this inequality with respect to time from 0 to t and invoking the fact that un(0)=um(0)=U0, we get
∥un(t)-um(t)∥B21(0,1)2≤C(hn+hm)+C∫0tsup0≤ξ≤t∥un(ξ)-um(ξ)∥B21(0,1)2dξ,whence
sup0≤s≤t∥un(s)-um(s)∥B21(0,1)2≤C(hn+hm)+C∫0tsup0≤ξ≤t∥un(ξ)-um(ξ)∥B21(0,1)2dξ.Accordingly, by Gronwall's lemma we obtain
sup0≤s≤t∥un(s)-um(s)∥B21(0,1)2≤C(hn+hm)exp(Ct),∀t∈[0,T],consequently
sup0≤s≤t∥un(s)-um(s)∥B21(0,1)≤Chn+hm
takes place for all n,m∈ℕ*. This implies that (un(t))n is a Cauchy sequence in the Banach space C(I,B21(0,1)), and hence it converges in the norm of this latter to some function u∈C(I,B21(0,1)). Besides, passing to the limit m→∞ in (5.27), we obtain the desired error estimate, which finishes the proof.

Now, we present some properties of the obtained solution.

The limit-function u from Proposition 5.2, possesses the following properties:

u∈C(I,B21(0,1))∩L∞(I,V)),

u is strongly differentiable a.e. in I and du/dt∈L∞(I,B21(0,1)),

ũn(t)→u(t) in B21(0,1) for all t∈I,

un(t), ũn(t)⇀u(t) in V for all t∈I,

(dun/dt)(t)⇀(du/dt)(t) in L2(I,B21(0,1)).

Proof.

On the basis of estimates (5.5) and (5.6), uniform convergence statement from Proposition 5.2, and the continuous embedding V↪B21(0,1), the assertions of the present theorem are a direct consequence of [2, Lemma 1.3.15].

Theorem 5.3.

Under Assumptions (H1)–(H3), (2.3) admits a unique weak solution, namely, the limit function u from Proposition 5.2, in the sense of Definition 2.3.

Proof.

We have to show that the limit function u satisfies all the conditions (i), (ii), (iii), and (iv) of Definition 2.3. Obviously, in light of the properties of the function u listed in Theorem 5.3, the first two conditions of Definition 2.3 are already seen. On the other hand, since un→u in C(I,V) as n→∞ and, by construction, un(0)=U0, it follows that u(0)=U0, so the initial condition is also fulfilled, that is, Definition 2.3(iii) takes place. It remains to see that the integral identity (2.21) is obeyed by u. For this, integrating (5.10) over (0,t) and using the fact that un(0)=U0, we get
(un(t)-U0,v)B21(0,1)+∫0t(ũn(τ),v)dτ=∫0t(k̃(τ,ũn(τ)),v)B21(0,1)dτ+∫0t(f̃n(τ),v)B21(0,1)dτ,consequently, after some rearrangements
(un(t)-U0,v)B21(0,1)+∫0t(ũn(τ),v)dτ=∫0t(∫0τa(τ-s)k(s,u(s))ds,v)B21(0,1)dτ+∫0t(f(τ),v)B21(0,1)dτ+∫0t(k̃(τ,ũn(τ))-∫0τa(τ-s)k(s,u(s))ds,v)B21(0,1)dτ+∫0t(f̃n(τ)-f(τ),v)B21(0,1)dτ.
Let ŝn:I→I and ŝn:I→I denote the functions
ŝn(t)={0,fort=0,tj-1n,inĨjn,s̃n(t)={0,fort=0,tjn,inĨjn.
To investigate the desired result, we prove some convergence statements. Using (5.2), (5.4), and (5.30) we have for allt∈(tj-1n,tjn]k̃(t,ũn(t))-∫0ta(t-s)k(s,u(s))ds=∫0tjn[a(tjn-ŝn(s))k(ŝn(s),ûn(s))-a(t-s)k(s,u(s))]ds+∫ttjna(t-s)k(s,u(s))ds.
Taking into account (5.5), (5.9), and assumptions (H1), (H2) it follows that
∥a(tjn-ŝn(s))k(ŝn(s),ûn(s))-a(t-s)k(s,u(s))∥B21(0,1)≤Chn.
Thanks to (5.31) and (5.32) we obtain
∥k̃(t,ũn(t))-∫0ta(t-s)k(s,u(s))ds∥B21(0,1)≤Chn.
On the other hand, in view of the assumed Lipschitz continuity of f, we have
∥f̃n(τ)-f(τ)∥B21(0,1)≤∥f(s̃n(τ))-f(τ)∥B21(0,1)≤l1hn.
Now, the sequences {(ũn(τ),v)}, {(f̃n(τ),v)B21(0,1)}, and {(k̃(τ,ũn(τ)),v)B21(0,1)} are uniformly bounded with respect to both τ and n, so the Lebesgue theorem of majorized convergence is applicable to (5.29). Thus, having in mind (5.7), (5.9), (5.33), and (5.34), we derive that
(u(t)-U0,v)B21(0,1)+∫0t(u(τ),v)dτ=∫0t(∫0τa(τ-s)k(s,u(s))ds,v)B21(0,1)dτ+∫0t(f(τ),v)B21(0,1)dτ
takes place for all v∈V and t∈[0,T]. Finally, differentiating (5.35) with respect to t, we get
(ddtu(t),v)B21(0,1)+(u(t),v)=(∫0ta(t-s)k(s,u(s))ds,v)B21(0,1)+(f(t),v)B21(0,1),a.e.t∈[0,T].The uniqueness may be argued in the usual manner. Indeed, exploiting an idea in [11], consider u1 and u2 two different solutions of (2.3), and define w=u1-u2 then, we have
(ddtw(t),v)B21(0,1)+(w(t),v)=(∫0ta(t-s)[k(s,u1(s))-k(s,u2(s))]ds,v)B21(0,1).Choosing v=w(t) as a test function, with the aid of Cauchy-Schwarz inequality and assumption (H1), we obtain
12ddt∥w(t)∥B21(0,1)2+∥w(t)∥2≤C∫0t[∥k(s,u1(s))-k(s,u2(s))∥B21(0,1)]ds∥w(t)∥B21(0,1).
Let ξ∈[0,p] such that
∥w(ξ)∥B21(0,1)=maxs∈[0,p]∥w(s)∥B21(0,1),
integrating (5.38) over (0,p), 0≤p≤T, using (5.39), and invokingassumption (H2), we get
∫0p[12ddt∥w(t)∥B21(0,1)2+∥w(t)∥2]dt≤Cp2∥w(ξ)∥B21(0,1)2,consequently, with the fact that w(0)=0∫0p[12ddt∥w(t)∥B21(0,1)2+∥w(t)∥2]dt≤Cp2∫0ξddt∥w(t)∥B21(0,1)2dt.
Choosing p as a constant verifying the condition
∃α∈ℕ,T=αp,Cp2≤12,we have, by virtue of (5.41)
∫0p12ddt∥w(t)∥B21(0,1)2dt+∫0p∥w(t)∥2dt≤∫0ξ12ddt∥w(t)∥B21(0,1)2dt,taking into account that ξ≤p, we obtain
∥w(t)∥=0,on[0,p].Following the same lines as for [0,p], we deduce that
∥w(t)∥=0,on[ip,(i+1)p],i=1,2,3,…,therefore, we derive w(t)≡0, on [0,T], then u1≡u2. This achives the proof.

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